Part 3

# Commensurable magnitudes

by Euclid

## Proposition 5

Commensurable magnitudes have to one another the ratio which (some) number (has) to (some) number.

E Let A and B be commensurable magnitudes. I say that A has to B the ratio which (some) number (has) to (some) number. For if A and B are commensurable (magnitudes) then some magnitude will measure them. Let it (so) measure (them), and let it be C. And as many times as C measures A, so many units let there be in D. And as many times as C measures B, so many units let there be in E. Therefore, since C measures A according to the units in D, and a unit also measures D according to the units in it, a unit thus measures the number D as many times as the magnitude C (measures) A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. Thus, inversely, as A (is) to C, so D (is) to a unit [Prop. 5.7 com.]. Again, since C measures B according to the units in E, and a unit also measures E according to the units in it, a unit thus measures E the same number of times that C (measures) B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And it was also shown that as A (is) to C, so D (is) to a unit. Thus, via equality, as A is to B, so the number D (is) to the (number) E [Prop. 5.22]. Thus, the commensurable magnitudes A and B have to one another the ratio which the number D (has) to the number E. (Which is) the very thing it was required to show. ers, rather than two number and two magnitudes. Proposition 6 If two magnitudes have to one another the ratio which (some) number (has) to (some) number then the magni- tudes will be commensurable. A Dr B Er F For let the two magnitudes A and B have to one an- other the ratio which the number D (has) to the number E. I say that the magnitudes A and B are commensu- rable.

For, as many units as there are in D, let A have been divided into so many equal (divisions). And let C be equal to one of them. And as many units as there are in E, let F be the sum of so many magnitudes equal to C. Therefore, since as many units as there are in D, so many magnitudes equal to C are also in A, therefore whichever part a unit is of D, C is also the same part of A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And a unit measures the number D. Thus, C also measures A. And since as C is to A, so a unit (is) to the [number] D, thus, inversely, as A (is) to C, so the number D (is) to a unit [Prop. 5.7 corr.]. Again, since as many units as there are in E, so many (magnitudes) equal to C are also in F, thus as C is to F, so a unit (is) to the [number] E [Def. 7.20]. And it was also shown that as A (is) to C, so D (is) to a unit. Thus, via equality, as A is to F, so D (is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B. And thus as A (is) to B, so (it) also is to F [Prop. 5.11]. Thus, A has the same ratio to each of B and F. Thus, B is equal to F [Prop. 5.9]. And C measures F. Thus, it also measures B. But, in fact, (it) also (measures) A. Thus, C measures (both) A and B. Thus, A is commensurable with B [Def. 10.1]. Thus, if two magnitudes… to one another, and so on

### Corollary

Thus, if there are 2 numbers, like D and E, and astraight-line, like A, then it is possible to contrive that as the number D (is) to the number E, so the straight-line (is) to (another) straight-line (ie., F). And if the mean proportion, (say) B, is taken of A and F, then as A is to F, so the (square) on A (will be) to the (square) on B. That is to say, as the first (is) to the third, so the (figure) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 com.]. But, as A (is) to F, so the number D is to the number E. Thus, it has also been contrived that as the number D (is) to the number E, so the (figure) on the straight-line A (is) to the (similar figure) on the straight-line B. (Which is) the very thing it was required to show.

## Proposition 7

Incommensurable magnitudes do not have to one an- other the ratio which (some) number (has) to (some) number. Let A and B be incommensurable magnitudes. I say that A does not have to B the ratio which (some) number (has) to (some) number.

For if A has to B the ratio which (some) number (has) to (some) number then A will be commensurable with B [Prop. 10.6]. But it is not. Thus, A does not have to B the ratio which (some) number (has) to (some) number. Thus, incommensurable numbers do not have to one another, and so on…. Proposition 8 If two magnitudes do not have to one another the ra- tio which (some) number (has) to (some) number then the magnitudes will be incommensurable. A+ B For let the two magnitudes A and B not have to one another the ratio which (some) number (has) to (some) number. I say that the magnitudes A and B are incom- mensurable. For if they are commensurable, A will have to B the ratio which (some) number (has) to (some) number [Prop. 10.5]. But it does not have (such a ratio). Thus, the magnitudes A and B are incommensurable. Thus, if two magnitudes… to one another, and so on

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